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Difference between revisions of "2004 IMO Problems/Problem 5"

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<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
 
<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
  
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  <math>\angle PLK=\frac12(\overarc{AD}+\overarc{CF}=\frac12(\overarc{CE}+\overarc{AB}=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
+
  <math>\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
 
  Since <math>AC//BF</math>, so <math>\triangle PBF</math> and <math>\triangle PDE</math> are isosceles triangles. So <math>P</math> is on the angle bisector oof <math>BF</math>, since <math>ABFC</math> is  
 
  Since <math>AC//BF</math>, so <math>\triangle PBF</math> and <math>\triangle PDE</math> are isosceles triangles. So <math>P</math> is on the angle bisector oof <math>BF</math>, since <math>ABFC</math> is  
 
  an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
 
  an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.

Revision as of 14:43, 8 February 2024

Problem

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.\]

Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

$\angle PBC=\angle DBA$, so $AD=CE$, and $DE//AC$. $\angle PDC=\angle BDA$, so $AB=CF$, and $AC//BF$. 

$\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL$, so $\triangle PKL$ is an isosceles triangle.
Since $AC//BF$, so $\triangle PBF$ and $\triangle PDE$ are isosceles triangles. So $P$ is on the angle bisector oof $BF$, since $ABFC$ is 
an isosceles trapezoid, so $P$ is also on the perpendicular bisector of $AC$. So $PA=PC$.


~szhangmath

See Also

2004 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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